Find the number of points of differentiability for the following function. If 
$$f(x)=\begin{cases}
    \cos x^3&;x\lt0\\
    \sin x^3 - |x^3-1|&;x\ge0
\end{cases}$$
then find the number of points where $g(x)=f(|x|) \text { is non differentiable.}$
 A: The question is asking about $f(|x|)$, by symmetry, since $g(x)$ is not differentiable at $x=1$, it is not differentiable at $x=-1$ as well. 
To show that it is not differentiable at $x=1$:
If it is differentiable at $x=1$, then the following limit exists.
\begin{align}\lim_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}&= \lim_{x \rightarrow 1} \frac{\sin x^3 - \sin 1-|x^3-1|}{x-1}\end{align}
since $\sin x^3$ is differentiable everywhre, the limit exists if and only if the following limit exists.
$$\lim_{x \rightarrow 1} \frac{|x^3-1|}{x-1}=\lim_{x \rightarrow 1} sign(x-1)|x^2+x+1|=\lim_{x \rightarrow 1}3sign(x-1)$$
Since $\lim_{x \rightarrow 1^+} sign(x-1) = 1 \neq -1 = \lim_{x \rightarrow 1^-} sign(x-1)$
The function is not differentiable at $x=1$.
Also, we have to verify that the function is differentiable at $x=0$.
$$f(|x|) = \begin{cases} \sin x^3-|x^3-1| & x \geq 0 \\ -\sin x^3 -|x^3+1| & x<0\end{cases}$$
When $|x|<1$, 
$$f(|x|) = \begin{cases} \sin x^3-1+x^3 & 0 < x < 1 \\ -\sin x^3 -x^3-1 & -1<x<0\end{cases}$$
$$\lim_{x \rightarrow 0}\frac{f(|x|)-f(0)}{x}=\lim_{x\rightarrow 0}\frac{sign(x)(\sin x^3+x^3)}{x}=0$$
